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There are several kind of summability , i accrossed differents conditions for applying for example Borel summation or laplace transform which let me mixed and confused , really i don't know if i have for example a complex valued function f or real valued presented as a formel series function to judge if it is summable or no , then my question here is :

Question: What is a sufficient condition for summability of formel power series ?

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    $\begingroup$ Trivial answer: absolute convergence is a sufficient condition for summability. $\endgroup$ Commented Dec 31, 2017 at 10:26
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    $\begingroup$ what about if it is also absolute divergent ? $\endgroup$
    – user119110
    Commented Dec 31, 2017 at 10:28

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Note The following strictly speaking does not answer the question but it may answer what the OP meant, i.e., under which conditions a formal power series defines a function.

Edit My memory didn't fail me. The following is taken almost verbatim from Reed and Simon Vol 4 see here

Definition 1 we say that a function $E\left(\lambda\right),$ analytic in a sectorial region $$\Omega=\left\{ z|0<\left|z\right|<B;\left|\textrm{arg}\left(z\right)\right|<\pi/2+\epsilon\right\} , $$ obeys a strong asymptotic condition and has $\sum_{n=0}^{\infty}a_{n}\lambda^{n}$ as strong asymptotic series (SAS) if there are positive constant $C$ and $\sigma$ such that $$ \left|E\left(\lambda\right)-\sum_{n=0}^{N}a_{n}\lambda^{n}\right|<C\sigma^{N+1}\left(N+1\right)!\left|\lambda\right|^{N+1} $$ for all $N$ and all $\lambda\in\Omega$.

Given the above one has:

Theorem A SAS defines a function in the sense that if two analitic functions $f,g$ have the same SAS then $f=g$.

Remark: Informally the coefficients must not grow too fast. In fact SAS implies $\left|a_{n}\right|\le C \sigma^{n}n!$.

There are series associated with simple examples for which $a_n$ behaves like $(kn)!$ with $k > 1$. Thus, a strong asymptotic condition cannot hold in such cases. However with a simple modification even this case may be treated. This suggests that we define:

Definition 2 we say that a function $E\left(\lambda\right),$ analytic in a sectorial region $$\Omega=\left\{ z|0<\left|z\right|<B;\left|\textrm{arg}\left(z\right)\right|<\pi/2+\epsilon\right\} ,$$ obeys a modified strong asymptotic condition of order $k$ and has $\sum_{n=0}^{\infty}a_{n}\lambda^{n}$ as an order $k$ strong asymptotic series if there are positive constant $C$ and $\sigma$ such that $$ \left|E\left(\lambda\right)-\sum_{n=0}^{N}a_{n}\lambda^{n}\right|<C\sigma^{N+1}\left [ k(N+1) \right]!\left|\lambda\right|^{N+1} $$ for all $N$ and all $\lambda\in\Omega$.

The above result extends to this case too. Namely

Theorem 2 If $f,g$ both have $\sum_{n=0}^{\infty}a_{n}\lambda^{n}$ as order $k$ strong asymptotic series, then $f=g$.

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  • $\begingroup$ @Icv , if you have the radius of convergence is 0 , can you say the coefficients not grow fast ? $\endgroup$
    – user119110
    Commented Dec 31, 2017 at 13:22
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    $\begingroup$ The definition seems needlessly complicated. For any positive $C$ and $\sigma$, there is a $C'$ so that $C\sigma^{N+1}\le C'(N+1)!$ for all $N$, so there's no need to have $\sigma$ in the definition. Ditto in your remark, the $k$ seems superfluous. Also, it would be easier to read if you displayed the definition of $\Omega$. $\endgroup$ Commented Dec 31, 2017 at 13:29
  • $\begingroup$ @JoeSilverman I agree. The reason is that such condition admits a generalization, which then looks more understandable. $\endgroup$
    – lcv
    Commented Dec 31, 2017 at 14:09
  • $\begingroup$ @JackWonder I'm not sure I understand. Yes of course. The point is to be able to define a function even for some series whose radius of convergence is zero. $\endgroup$
    – lcv
    Commented Dec 31, 2017 at 14:11
  • $\begingroup$ @JoeSilverman I added the generalization and found the reference. Basically these are applications/extensions of Carleman's theorem. $\endgroup$
    – lcv
    Commented Dec 31, 2017 at 14:33